Restriction Enzymes

Correct answer: A. In complete digestion fragments equal cuts

In circular DNA after complete digestion fragments equal the number of cut sites. This is useful in plasmid questions.

Correct answer: A. many enzymes recognize palindromic DNA sequences / Many enzymes recognize palindromic DNA sequences

Many restriction enzymes recognize palindromic DNA sequences as recognition sites. This is a basic concept.

Correct answer: A. Restriction enzymes are bacteria-linked DNA cutting tools

The source of restriction enzymes is bacteria and the function is DNA cutting. This is the most important easy-level summary.

Correct answer: B. Five

In linear DNA, the restriction enzyme makes one cut at each restriction site. Therefore, with 4 cuts, the number of fragments is one more than the number of cuts: \(4+1=5\). Three fragments is incorrect because linear DNA does not produce fewer fragments than its cuts in complete digestion. Exam tip: For linear DNA, remember: fragments = cuts + 1.

Correct answer: C. Four

The direct answer is C: four fragments. For a circular DNA molecule, if there are four distinct restriction sites and digestion is complete, the four cuts divide the circular molecule into four separate arcs. Each arc becomes one DNA fragment, so the number of fragments equals the number of cuts. Option C is correct. Option A, three, would be expected for a linear DNA molecule with four internal cuts, because a linear molecule gives cuts plus one fragments. Option B, five, is also the linear-DNA rule applied incorrectly or an overcount. Option D, one, ignores the fact that four cuts separate the circle into multiple pieces; one fragment would describe an uncut plasmid, not complete digestion. This answer assumes all four sites are distinct, accessible, and actually cut by the enzyme. The useful rule is: circular DNA with n cuts gives n fragments; linear DNA with n cuts gives n+1 fragments. Memory cue: close the circle, so the last cut joins back to the first; four cuts make four pieces.

Correct answer: D. Two

At one restriction site, the enzyme makes one cut in linear DNA. This produces two separate DNA pieces, so the correct answer is two. Three fragments would generally require two cuts in linear DNA. Exam tip: for linear DNA, number of fragments = number of cuts + 1.

Correct answer: A. One linear DNA molecule

One cut in a circular plasmid makes one linear DNA molecule. This is a basic rule of plasmid digestion.

Correct answer: B. To separate DNA fragments by size

After restriction digestion, one DNA molecule may be converted into fragments of different lengths. Agarose gel electrophoresis separates these charged DNA fragments mainly according to size: when an electric field is applied, DNA moves toward the positive electrode, and smaller fragments generally migrate through the gel matrix faster and farther than larger fragments. The resulting bands reveal the restriction pattern and allow comparison or approximate size estimation with a DNA ladder. Therefore option B is correct. Gel electrophoresis does not multiply the enzyme, convert DNA into sugar, or dissolve protein; those statements confuse separation with unrelated biochemical processes.

Correct answer: C. Moves farther

The direct answer is option C: a smaller DNA fragment generally moves farther through a gel than a larger fragment. In gel electrophoresis, an electric field pulls negatively charged DNA through tiny spaces in the gel. The gel acts like a sieve. Small fragments pass through these spaces more easily and therefore travel a greater distance in the same time, while large fragments face more resistance. Option A is wrong because it describes the usual behavior of a larger fragment, not a smaller one. Option B is wrong because DNA fragments do not normally remain at exactly the loading position. Option C is correct because size-based separation makes smaller pieces migrate farther. Option D is wrong because DNA does not become protein during gel electrophoresis. Exam cue: in a DNA gel, farther usually means smaller, provided the conditions are the same.

Correct answer: A. EcoRI-GAATTC

EcoRI is a restriction endonuclease whose well-known recognition sequence is 5′-GAATTC-3′. It cuts within this palindromic site and produces cohesive, or sticky, DNA ends. HaeIII is associated with GGCC, not GAATTC, while BamHI recognizes GGATCC rather than AAGCTT. The sequence AAGCTT is associated with HindIII, so option D is also mismatched. Thus option A is the only correct enzyme–recognition-sequence pairing. Correct matching is important because the choice of restriction enzyme determines where a vector or insert will be cut and whether their ends will be compatible for ligation.

Correct answer: A. BamHI — GGATCC

BamHI is a restriction enzyme that recognizes and cuts the palindromic recognition sequence GGATCC. Therefore, option A is correct. GAATTC is recognized by EcoRI, AAGCTT by HindIII, and GGCC by HaeIII. Exam tip: Memorize common restriction enzymes together with their recognition sequences as pairs.

Correct answer: A. HindIII recognition sequence: AAGCTT

HindIII is a restriction enzyme that recognises and cuts the palindromic recognition sequence AAGCTT. Therefore, option A is correct. GGATCC is recognised by BamHI, whereas GAATTC is recognised by EcoRI; CCGG is recognised by enzymes such as HpaII/MspI. Exam tip: Memorise common restriction enzyme names along with their recognition sequences.

Correct answer: A. GGCC and blunt ends

Direct answer: A, GGCC and blunt ends. HaeIII is a restriction endonuclease that recognizes the palindromic sequence 5′-GGCC-3′ and cuts in a way that leaves blunt ends, with no single-stranded overhang. This is the exact fact needed. Option A is correct because it gives both the recognition sequence and the type of DNA ends. Option B is wrong: GAATTC is the recognition sequence of EcoRI, which produces sticky ends, not the standard HaeIII example. Option C is wrong: AAGCTT is associated with HindIII, and a hormone is unrelated to restriction-site description. Option D is wrong: GGATCC is associated with BamHI, while starch is not a product or feature of HaeIII recognition. Blunt ends can be joined, but they generally lack the easy complementary pairing found in sticky ends, so ligation may be less efficient. Exam cue: HaeIII means GGCC and blunt; EcoRI means GAATTC and sticky.

Correct answer: B. neoschizomer

Neoschizomers can recognize the same sequence but may cut at different positions. This is a terminology-based question.

Correct answer: B. isoschizomers

Isoschizomers are different enzymes that recognize the same recognition sequence. Remember the term with its meaning.

Correct answer: A. DNA recognition sequence

A restriction enzyme recognizes its specific DNA recognition sequence. This defines its specificity.

Correct answer: A. The ends are complementary

The single-stranded overhangs of complementary sticky ends can base-pair with each other. This holds the two DNA fragments in the correct alignment, allowing DNA ligase to form permanent phosphodiester bonds in the sugar-phosphate backbone. Methylation mainly affects cutting by restriction enzymes; it is not the essential condition that makes ligation easier. Exam tip: complementary base pairing makes sticky-end ligation more efficient.

Correct answer: A. Overhang is absent

The direct answer is option A: blunt-end joining can be comparatively difficult when the overhang is absent. A sticky end has a short single-stranded overhang that can pair with a complementary overhang on another DNA molecule. This temporary base pairing holds the molecules together and helps ligase find the correct position. Blunt ends have no such overhang, so there is no initial complementary pairing; ligase can still join them, but the reaction is usually less efficient and may need suitable conditions. Option A is correct because absence of an overhang describes a blunt end. Option B is wrong because a sticky overhang generally makes joining easier. Option C is wrong because active ligase supports joining rather than causing difficulty. Option D is wrong because a correct buffer also supports the reaction. Memory cue: sticky ends stick first; blunt ends need more help.

Correct answer: A. Found in bacteria and cut DNA

Restriction enzymes are bacterial-origin DNA cutting tools. This easy-level summary is most important.

Correct answer: D. They mainly synthesize glucose

Restriction enzymes are bacterial endonucleases that recognize specific DNA sequences and cut DNA at or near those sites. This property makes them useful in recombinant DNA technology, and they also help protect bacteria against foreign DNA such as bacteriophage DNA. They do not synthesize glucose, so option D is the incorrect statement. Exam tip: remember restriction enzymes as the molecular scissors of DNA technology.

Correct answer: A. gene cloning

In gene cloning restriction enzymes help cut vector and insert DNA. This is a biotechnology application.

Correct answer: C. starch digestion

Restriction enzymes are useful in DNA analysis and recombinant DNA. Enzymes like amylase are used for starch digestion.

Correct answer: A. multiple cloning site

A multiple cloning site has many restriction sites. It provides flexibility in insert DNA cloning.

Correct answer: A. Low DNA or weak staining

Low DNA or weak staining can make small bands hard to see. Remember sample amount in practical interpretation.

Correct answer: A. partial digestion

The direct answer is option A: an extra uncut band can indicate partial digestion. In restriction digestion, an enzyme should cut DNA at its specific recognition sites. If the reaction is incomplete, some DNA molecules are cut while others remain uncut. Gel electrophoresis then shows both the expected digestion fragments and a band representing uncut DNA, so an extra band may suggest partial digestion. Option A is correct, although an uncut band can have other technical explanations, such as unsuitable enzyme conditions or resistant DNA forms. Option B is wrong because complete digestion should not normally leave an uncut band of the original material. Option C is wrong because an uncut band does not prove cloning, much less perfect cloning. Option D is wrong because a visible DNA band means DNA is present. Exam cue: cut bands plus an original-looking uncut band often means incomplete digestion.